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Part 5.
The chemistry of
ALDEHYDES and KETONES
Nucleophilic addition reactions
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Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study
Notes for UK KS5 A/AS GCE IB advanced level organic chemistry students US
K12 grade 11 grade 12 organic chemistry
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notes on the chemistry of aldehydes and ketones
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Part 6.4
Structure and reactivity of aldehydes and ketones - the nucleophilic
addition of hydrogen cyanide and the hydrolysis of the nitriles formed and
stereochemistry of nucleophilic addition to carbonyl compounds
Sub-index for this page
5.4.1.
Structure and reactivity of aldehydes
and ketones - nucleophilic attack
5.4.2
The reaction between
hydrogen cyanide and aldehydes or ketones to form a nitrile
5.4.3
The mechanism of nucleophilic addition
of hydrogen cyanide to aldehydes and ketones
5.4.4.
Addition mechanisms - comparing aldehydes/ketones with alkenes
5.4.5
The hydrolysis of the nitrile to a carboxylic acid
5.4.6
The stereochemistry of
nucleophilic addition to aldehydes and ketones - implications
- comparing the reactivity modes of alkenes compared to aldehydes/ketones
5.4.1 Structure and reactivity of aldehydes
and ketones
Structure reminders
Note: An aryl group means a benzene
ring C6H5- or a substituted benzene ring e.g. CH3C6H5-.
Above is a diagram of the sigma and pi
bonds of the carbonyl group formed by the overlap of the 2s/2p orbitals of a
carbon and oxygen atom.
The pi electron clouds lie above and
below the plane of the >C-O sigma bond system, one electron in each pi
orbital., formed specifically from the overlap of 2p orbitals
The bond is polar because of the difference in
electronegativities of the atoms forming the carbonyl bond, on the
Pauling scale carbon is 2.5 and oxygen 3.5.
This produces a shift in the electron
clouds towards the more electronegative oxygen atom giving aldehydes and
ketones a partially positive carbon atom susceptible to nucleophilic
attack by an electron pair donor - a nucleophile e.g. the cyanide ion
-:C≡N
(more details further down the page).
The next diagram illustrates the mode of nucleophilic
attack on the carbonyl group of aldehydes and ketones.
The geometry of the functional group
of aldehydes and ketones
There is a trigonal planar
arrangement of bonds around the carbon atom of the carbonyl group (>C=O)
because there are three groups of electrons involved.
There are two C-R σ bonds
between the carbon and the H/alky/aryl group and
the σ
plus π of the carbon - oxygen bond of the
carbonyl C=O group which spread out in an angular manner of 120o to minimise
repulsion.
The planarity of this bond arrangement means that
there is a 50:50 chance of the nucleophile attacking on either side of
carbon - this has consequences if the product is an optical R/S isomer.
To envisage this, consider the right-hand side of the
diagram and imagine the C=O in the plane of the screen and the 'black'
bond ▲ pointing directly towards you and the 'grey' bond▼
pointing directly away from you.
This is further discussed with extra mechanism
diagrams in
section 5.4.6.
The reactive nature of the carbonyl
group in aldehydes and ketones
Aldehydes and ketones
readily undergo nucleophilic attack because of the highly
polar carbonyl bond >Cδ+=Oδ–
caused by the big difference in the electronegativity between carbon (2.5) and
oxygen (3.5). The π electrons
The more electronegative
oxygen pulls the bonding electron cloud towards itself producing the
highly bond which is the basis for most of the chemical reactions of
aldehydes and ketones. An electron pair donating nucleophile (neutral
:Nuc or negative
:Nuc-), will
therefore attack the 'positive carbon' (Cδ+)
to form a C–Nuc bond.
The relative
reactivity of some aldehydes and ketones
The methyl group (-CH3)
has a small electron cloud releasing effect (+ inductive effect →) that
partially reduces the effect of the oxygen on creating the delta plus
carbon atom.
Methanal has no methyl group,
ethanal has one methyl group and propanone has two methyl groups,
hence the reactivity trend:
HCHO > CH3→CHO
> CH3→CO←CH3
From left to right, the
delta plus isn't quite as positive!
Generally speaking,
aldehydes are more reactive than ketones.
Comparison of the mode of reactivity
of alkenes compared to aldehydes and ketones.
A comparison of
electrophilic addition to alkenes with nucleophilic addition to
aldehydes/ketones
TOP OF PAGE
5.4.2 The reaction between
hydrogen cyanide and aldehydes or ketones
The
product of the nucleophilic addition of hydrogen cyanide is a hydroxynitrile (a
cyanohydrin).
The reaction is equivalent to adding H-CN
across the C=O bond, to give a N-C-O-H bonding situation.
The addition begins with the initial addition of a
cyanide ion (see details of mechanism in
section 5.4.3).
So, the reagent and reaction conditions must be just
right.
Hydrogen cyanide is a very weak acid (Ka = 5 x 10-10
mol dm-3), and, on its own at equilibrium, it produces very few
cyanide ions:
(i) HCN + H2O
H3O+ + :CN-
Therefore a base (alkali) must be present to raise the pH
>7.
The presence of a strong base (e.g. hydroxide ion)
generates a sufficiently high concentration of cyanide ions to allow the
addition of HCN to proceed efficiently.
(ii) HCN + OH-
H2O + :CN-
If the pH is too low, there are insufficient cyanide ions
for the reaction to proceed quickly.
The lower the pH, the greater the danger from toxic hydrogen cyanide
gas.
In practice, a solution of potassium cyanide (KCN) is
used, buffered to about pH ~8.
KCN is the salt of a strong base and
a very weak acid, and is naturally alkaline by hydrolysis (the reverse
of reaction (ii) above), so providing a higher concentration of cyanide
ions than hydrogen cyanide. To
get the right pH, a little dilute sulfuric acid is added to a solution
of sodium/potassium cyanide.
Using pure HCN solution, the reaction takes weeks, add a drop of NaOH
and it goes in hours, so the base (alkali) has quite a catalytic effect.
Examples of
nucleophilic addition of hydrogen cyanide to aldehydes and ketones
to give hydroxynitriles
(a)
+ HCN ===>
ethanal + hydrogen
cyanide ===> 2-hydroxypropanenitrile
(b)
+ HCN ===>
propanone + hydrogen
cyanide ===> 2-hydroxy-2-methylpropanenitrile
(c)
+ HCN ===>
butanone + hydrogen
cyanide ===> 2-hydroxy-2-methylbutanenitrile
Need some bigger skeletal formulae equations
TOP OF PAGE
and sub-index
5.4.3 The
mechanism of nucleophilic addition
of hydrogen cyanide
to aldehydes and ketones
The reagent and reaction conditions were
discussed above in section
6.4.2.
mechanism 7 –
nucleophilic addition of cyanide ion to an aldehyde or ketone
[mechanism
7 above] The >Cδ+=Oδ–
bond is highly polarised
because of the great difference in electronegativity between carbon (2.1)
and oxygen (3.5).
Step
(1) The
nucleophilic electron pair donating cyanide ion attacks the
positive carbon of the polarised C=O bond, forming a C–C bond.
The cyanide ion is the nucleophile - donating an
electron pair to a partially positive carbon atom.
The lone pair of electrons on the
carbon of the cyanide ion,
:CN-,
forms a C-C bond
with the
δ+ carbon of the carbonyl group.
Then the electron shift as the
π
electron bond pair of the
original C=O bond moves onto the oxygen to give it a whole negative
charge.
Step
(2) The
negative ion (anion) intermediate
formed, RR'C(CN)O–, is a strong conjugate base and will
abstract a proton from water or hydrogen cyanide to give the hydroxynitrile product and
a hydroxide ion.
For step (2) you can also write:
RR'C(CN)O–
+ HCN ===> RR'C(CN)OH + CN–
but I'm not sure which dominates the
proton donation to form the hydroxy-nitrile?
Mechanism diagram mechanism 77a shows the
nucleophilic addition of hydrogen cyanide ion to the aldehyde ethanal.
The initial attacking nucleophile
(electron pair donor) on the ethanal molecule is the cyanide ion.
The intermediate anion abstracts a proton
from water (or hydrogen cyanide) to form the product
2-hydroxypropanenitrile.
Mechanism diagram mechanism 77b shows the
nucleophilic addition of hydrogen cyanide ion to the ketone propanone.
The initial attacking nucleophile
(electron pair donor) on the propanone molecule is the cyanide ion.
The intermediate anion abstracts a proton
from water (or hydrogen cyanide) to form the product 2-hydroxy-2-methyl
propanenitrile.
TOP OF PAGE
and sub-index
5.4.4 FURTHER
COMMENTS
- comparing the reactivity modes of alkenes compared to aldehydes/ketones
A
comparison
of the mode of addition alkenes compared to carbonyl compounds
Why do alkenes
react by electrophilic addition and carbonyl compounds by
nucleophilic addition?
Both alkenes (C=C) and carbonyl
compounds (C=O) contain
π
bonds, the
π
electron cloud is above and below
the plane of the C-C or C-O σ bond.
The C=C bond is
non-polar, but the C=O bond is,
δ+C=Oδ–,
due to the difference in electronegativity (carbon 2.5 and oxygen 3.5).
Alkene reactivity:
Electron pair donating
nucleophiles, especially if negative (e.g. X– or OH–)
will tend to be repelled by the high electron density of the
π
bond in alkenes.
In alkenes,
the electron pair ('rich') donating double bond, is much more
likely to react with an electron pair accepting electrophile
(Lewis acid) like a positive ion.
Alkenes are also less susceptible
to nucleophilic attack because the C=C bond is non-polar.
Aldehyde and ketone reactivity:
Although electron pair donating
nucleophiles tend to be repelled by the high electron density of the
π
bond, in
carbonyl compounds, the 'distorted' highly polar >Cδ+=Oδ–
bond, will be susceptible to attack at the positive carbon (δ+C) by
electron pair donating nucleophiles.
The Oδ–
oxygen atom is not attacked by electrophiles such as HBr or Br2
because the oxygen atom is too electronegative - too reluctant to act as
a lone pair donor.
TOP OF PAGE
and sub-index
 5.4.5
The hydrolysis of the nitrile to a carboxylic acid
The general equation for the hydrolysis
of a hydroxy nitrile to a hydroxy-carboxylic acid.
R2C(OH)CN
+ 2H2O
===> R2C(OH)COOH
+ NH3
The nitrile is converted to a carboxylic acid by hydrolysis.
The reaction is slow, so it is speeded up by heating the
nitrile with the hydrolysis reagent (e.g. dilute sulfuric acid or aqueous
sodium hydroxide) under reflux.
(1) Hydrolysis
equation with water - very slow, even under reflux.
(a)
+ 2H2O ===>
 + NH3
2-hydroxypropanenitrile +
water ===> 2-hydroxypropanoic acid + ammonia
The products can also be expressed as the ammonium salt
(b)
+ 2H2O ===>
 + NH3
2-hydroxy-2-methylpropanenitrile +
water ===> 2-hydroxy-2-methylpropanoic acid + ammonia
The products can also be expressed as the ammonium salt
(c)
+ 2H2O ===>
 + NH3
2-hydroxy-2-methylbutanenitrile
+ water ===> 2-hydroxy-2-methylbutanoic acid +
ammonia
The products can also be expressed as the ammonium salt
(2)
Hydrolysis
equation with acid - good yield under reflux conditions
(a)
+ H+ + 2H2O ===>
+ NH4+
2-hydroxypropanenitrile +
hydrogen ion ===> 2-hydroxypropanoic acid + ammonium
ion
Note that this produces the free acid.
(b)
+ H+ + 2H2O ===>
+ NH4+
2-hydroxy-2-methylpropanenitrile + hydrogen ion ===> 2-hydroxy-2-methylpropanoic acid + ammonium
ion. Note that this reaction produces the free acid.
(c)
+
H+ + H2O ===>
+
NH4+
2-hydroxy-2-methylbutanenitrile
+ water ===> 2-hydroxy-2-methylbutanoic acid +
ammonia
Note that this produces the free acid.
(3)
Hydrolysis
with alkali - good yield under reflux conditions
(a)
+ OH- + H2O ==>
 + NH3
2-hydroxypropanenitrile +
hydroxide in ===> 2-hydroxypropanoate ion + ammonia
Note that this produces the salt of the acid and the ammonia
would be boiled off under reflux conditions.
e.g. from sodium hydroxide you would get sodium
2-hydroxypropanoate.
You add dilute mineral acid (sulfuric or hydrochloric) to
free the acid - strong acid displacing a weak one.
(b)
+ OH- + H2O ==>
 + NH3
2-hydroxy-2-methylpropanenitrile + hydroxide in ===> 2-hydroxy-2-methylpropanoate
ion + ammonia
Note that this produces the salt of the acid and the ammonia
would be boiled off under reflux conditions.
e.g. from sodium hydroxide you would get sodium
2-hydroxy-2-methylpropanoate.
You add dilute mineral acid (sulfuric or hydrochloric) to
free the acid - strong acid displacing a weak one.
(c)
+
OH- + H2O ==>
 + NH3
2-hydroxy-2-methylbutanenitrile
+ water ===> 2-hydroxy-2-methylbutanoate ion +
ammonia
Note that this produces the salt of the acid and the ammonia
would be boiled off under reflux conditions.
e.g. from potassium hydroxide you would get potassium
2-hydroxy-2-methylbutanoate
You add dilute mineral acid (sulfuric or hydrochloric) to
free the acid - strong acid displacing a weak one.
The acid, here 2-hydroxy-2-methylbutanoic acid, is set free by
the addition of a strong mineral acid e.g. dilute sulfuric acid.
CH3CH2C(CH3)(OH)COO-(aq) + H+(aq)
===> CH3CH2C(CH3)(OH)COOH
Note on R/S isomers
If the nitrile is derived from aldehydes from ethanal (CH3CHO)
onwards and all unsymmetrical ketones (e.g. butanone CH3COCH2CH3),
R/S isomers are formed on equal probability basis - a racemic mixture
(optically inactive racemate). In other words, if the resulting nitriles from
these carbonyl compounds are hydrolysed, you will also get a racemic (50 :
50)mixture of the mirror image forms of the hydroxynitrile (see section
5.4.6 for more details).
Apart from hydrolysis of nitriles to carboxylic acids, they
can also be reduced to hydroxylamines.
For details see
section Part 5.5 for the details of
reduction reactions.
TOP OF PAGE
and sub-index
5.4.6
The stereochemistry of nucleophilic addition to aldehydes and ketones
Implications if the product can
exist as R/S stereoisomers
In this nucleophilic
addition reaction, at the functional group centre of the reaction
(>C=O), you change from an unsaturated trigonal planar situation to a
saturated tetrahedral bond network about the carbon atom.
This carbon
atom is, in most cases a chiral carbon and the product therefore can
exhibit optical
R/S isomerism.
However the product is usually a 50:50 mixture of the
enantiomers (non–superimposable mirror–image forms) i.e. an optically
inactive racemic mixture.
Why is the product an
optically inactive racemate even if the product is an asymmetric
molecule with a chiral carbon?
The reason can be clearly
argued by considering the right-hand of the diagram
above.
The nucleophile attacks the carbon of the polarised carbonyl
group (R2Cδ+=Oδ–)
in a trigonal planar bonding situation which changes to a tetrahedral on
formation of the C–Nucleophile bond.
Consider the right-hand side of the
diagram and imagine the C=O in the plane of the screen and the 'black'
bond ▲ pointing directly towards you and the 'grey' bond▼
pointing directly away from you. Quite simply, there is a 50:50
chance of which side of the carbonyl group the nucleophile attacks and
therefore a 50:50 chance of which optical isomer is formed as the
configuration about the carbon atom changes.
Apart from explaining the
formation of a racemic mixture, you can also argue, in turn, that the
lack of optical activity in the product is itself evidence for an
initial attack of the nucleophile at the carbon of the carbonyl group
and you might reasonably expect a 2nd order rate expression.
rate = k2[aldehyde/ketone][CN–]
Though I don't know if the
kinetics are actually this simple for what seems to be an initial bimolecular rate
determining step mechanism!
Picturing this 'theoretically' as 50:50 in terms of the whole reaction mechanism
products
Again, imagine the C=O in the plane of the screen and the 'black'
bond ▲ pointing directly towards you and the 'grey' bond▼
pointing directly away from you and the N≡C-C-O bond in the plane of
the screen. If R = R', the
molecule has a plane of symmetry and cannot exhibit R/S stereoisomerism -
you can't get two non-superimposable mirror image forms of the molecule.
This is the case if the aldehyde is methanal (HCHO) or a
symmetrical ketone (e.g. propanone CH3COCH3).
However, for all other aldehydes from ethanal (CH3CHO)
onwards and all unsymmetrical ketones (e.g. butanone CH3COCH2CH3),
R/S isomers are formed on equal probability basis giving a racemic mixture
(optically inactive racemate). Another consequence of this is if the resulting
nitrile is hydrolysed, you will also get a racemic mixture of the
hydroxynitriles (see section
5.4.5).
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